我想将"01110"替换成"01111",要求是取出"01110"中最后那一位的"0",将其替换成"1",结果是"01111",高手请指教
情况1：
如果确定在最后这个位置，
String old = "01110";
String n = old.substring(0, old.length() - 1) + "1";
情况2：
如果不在最后位置
String old = "11110111";
int pos = old.lastIndexOf("0");
String n = old;
if(pos > -1){
n = old.substring(0, pos) + "1" + (pos < old.length() - 1?old.substring(pos + 1) : "");
}
你觉得哪里有难度??
String a = "01110";
a = "01111";
a.substring(0, index) + "1" + a.substring(index + 1)
1、
[code="java"]
int index = 1;//第几个0,start at 0
String old = "011101110";
int i = -1, pos = -2;
while(i < index && pos != -1){
pos = old.indexOf("0", pos == -2 ? 0 : pos + 1);
if(pos > -1){
++i;
}
}
String n = "not found";
if(i == index){
n = old.substring(0, pos) + 1 + (pos < old.length() -1 ? old.substring(pos + 1, old.length()) : "")
}
[/code]
public String relay(String s, int index){
long old=Long.paseLong(s);
return ""+(1L << index | old);
}
public String XX(String s, int start, int end){
return s.subString(0,start-1)+~Long.parseLong(s.subString(start,end))+s.subString(end));
}
修改下mkey的,start，end就不判定边界和顺寻了
[code="java"]String sub = s.substring(start, end);
String a = Long.toBinaryString(~Long.parseLong(sub, 2));
return (start>0?s.substring(0, start-1) : "") + a.substring(a.length() + start - end) + s.substring(end);
[/code]
楼主学过编程吗
你觉着那段代码逻辑合适就自己整个函数，有点同意qinglangee，要是这样的话，建议系统学习下
以下是一个方法
[code="java"]public String genXX(String old, int index){
//int index = 1;//第几个0,start at 0
//String old = "011101110";
int i = -1, pos = -2;
while(i < index && pos != -1){
pos = old.indexOf("0", pos == -2 ? 0 : pos + 1);
if(pos > -1){
++i;
}
}
String n = "not found";
if(i == index){
n = old.substring(0, pos) + 1 + (pos < old.length() -1 ? old.substring(pos + 1, old.length()) : "")
}
}[/code]
上面忘了加return
[code="java"]public String genXX(String old, int index){
//int index = 1;//第几个0,start at 0
//String old = "011101110";
int i = -1, pos = -2;
while(i < index && pos != -1){
pos = old.indexOf("0", pos == -2 ? 0 : pos + 1);
if(pos > -1){
++i;
}
}
String n = "not found";
if(i == index){
n = old.substring(0, pos) + 1 + (pos < old.length() -1 ? old.substring(pos + 1, old.length()) : "")
}
return n;
} [/code]